12—Tensors 303
Alternating Tensor
A curious (and very useful) result about antisymmetric tensors is that in three dimensions there is, up
to a factor, exactly one totally antisymmetric third rank tensor; it is called the “alternating tensor.”
So, if you take any two such tensors,ΛandΛ′, then one must be a multiple of the other. (The same
holds true for thenthrank totally antisymmetric tensor inndimensions.)
Proof: Consider the functionΛ−αΛ′whereαis a scalar. Pick any three independent vectors
~v 10 ,~v 20 ,~v 30 as long asΛ′on this set is non-zero. Let
α=
Λ(~v 10 , ~v 20 , ~v 30 )
Λ′(~v 10 , ~v 20 , ~v 30 )
(12.24)
(IfΛ′gives zero for every set of~vs then it’s a trivial tensor, zero.) This choice ofαguarantees that
Λ−αΛ′will vanish for at least one set of values of the arguments. Now take a general set of three
vectors~v 1 ,~v 2 , and~v 3 and ask for the effect ofΛ−αΛ′on them.~v 1 ,~v 2 , and~v 3 can be expressed as
linear combinations of the original~v 10 ,~v 20 , and~v 30. Do so. Substitute intoΛ−αΛ′, use linearity and
notice that all possible terms give zero.
The above argument is unchanged in a higher number of dimensions. It is also easy to see that
you cannot have a totally antisymmetric tensor of rankn+ 1inndimensions. In this case, one of the
n+ 1variables would have to be a linear combination of the othern. Use linearity, and note that when
any two variables equal each other, antisymmetry forces the result to be zero. These observations imply
that the function must vanish identically. See also problem12.17. If this sounds familiar, look back at
section7.7.
12.4 Birefringence
It’s time to stop and present a serious calculation using tensors, one that leads to an interesting and
not at all obvious result. This development assumes that you’ve studied Maxwell’s electromagnetic
field equations and are comfortable with vector calculus. If not then you will find this section obscure,
maybe best left for another day. The particular problem that I will examine is how light passes through
a transparent material, especially a crystal that has different electrical properties in different directions.
The common and best known example of such a crystal is Iceland spar, a form of calcite (CaCO 3 ).
∇×E~=−
∂B~
∂t
∇×B~=μ 0 ~j+μ 0 0
∂E~
∂t
~j=∂
P~
∂t
P~=α(E~) (12.25)
E~ andB~are the electric and magnetic fields.P~is the polarization of the medium, the electric dipole
moment density.αis the polarizability tensor, relating the amount of distortion of the charges in the
matter to the applied electric field. The current density~jappears because a time-varying electric field
will cause a time-varying movement of the charges in the surrounding medium — that’s a current.
Take the curl of the first equation and the time derivative of the second.
∇×∇×E~=−
∂
∂t
∇×B~ ∇×
∂B~
∂t
=μ 0
∂~j
∂t
+μ 0 0
∂^2 E~
∂t^2
The two expressions involvingB~are the same, so eliminateB~.
∇×∇×E~=−μ 0
∂~j
∂t
−μ 0 0
∂^2 E~
∂t^2
=−μ 0
∂^2 α
(~
E
)
∂t^2
−μ 0 0
∂^2 E~
∂t^2
I make the assumption thatαis time independent, so this is
∇×∇×E~=∇
(
∇.E~
)
−∇^2 E~=−μ 0 α
(
∂^2 E~
∂t^2
)
−μ 0 0
∂^2 E~
∂t^2
